Solve frac{2-2n^{2}}{(2+2n^2)^2}=1/9+2-2n^2/2+2n^2

$$\frac { 2 - 2 n ^ { 2 } } { ( 2 + 2 n ^ { 2 } ) ^ { 2 } } = \frac { 1 } { 9 } + \frac { 2 - 2 n ^ { 2 } } { 2 + 2 n ^ { 2 } }$$

Solve for n

$$n = -\frac{\sqrt{6 \sqrt{97} + 26}}{8} \approx -1.153074334$$

$$n = \frac{\sqrt{6 \sqrt{97} + 26}}{8} \approx 1.153074334$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 36\left(n^{2}+1\right)^{2}, the least common multiple of \left(2+2n^{2}\right)^{2},9,2+2n^{2}.

$$9\left(2-2n^{2}\right)=36\left(n^{2}+1\right)^{2}\times \frac{1}{9}+\left(18n^{2}+18\right)\left(2-2n^{2}\right)$$

Use the distributive property to multiply 9 by 2-2n^{2}.

$$18-18n^{2}=36\left(n^{2}+1\right)^{2}\times \frac{1}{9}+\left(18n^{2}+18\right)\left(2-2n^{2}\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(n^{2}+1\right)^{2}.

$$18-18n^{2}=36\left(\left(n^{2}\right)^{2}+2n^{2}+1\right)\times \frac{1}{9}+\left(18n^{2}+18\right)\left(2-2n^{2}\right)$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$18-18n^{2}=36\left(n^{4}+2n^{2}+1\right)\times \frac{1}{9}+\left(18n^{2}+18\right)\left(2-2n^{2}\right)$$

Multiply 36 and \frac{1}{9} to get 4.

$$18-18n^{2}=4\left(n^{4}+2n^{2}+1\right)+\left(18n^{2}+18\right)\left(2-2n^{2}\right)$$

Use the distributive property to multiply 4 by n^{4}+2n^{2}+1.

$$18-18n^{2}=4n^{4}+8n^{2}+4+\left(18n^{2}+18\right)\left(2-2n^{2}\right)$$

Use the distributive property to multiply 18n^{2}+18 by 2-2n^{2} and combine like terms.

$$18-18n^{2}=4n^{4}+8n^{2}+4-36n^{4}+36$$

Combine 4n^{4} and -36n^{4} to get -32n^{4}.

$$18-18n^{2}=-32n^{4}+8n^{2}+4+36$$

Add 4 and 36 to get 40.

$$18-18n^{2}=-32n^{4}+8n^{2}+40$$

Add 32n^{4} to both sides.

$$18-18n^{2}+32n^{4}=8n^{2}+40$$

Subtract 8n^{2} from both sides.

$$18-18n^{2}+32n^{4}-8n^{2}=40$$

Combine -18n^{2} and -8n^{2} to get -26n^{2}.

$$18-26n^{2}+32n^{4}=40$$

Subtract 40 from both sides.

$$18-26n^{2}+32n^{4}-40=0$$

Subtract 40 from 18 to get -22.

$$-22-26n^{2}+32n^{4}=0$$

Substitute t for n^{2}.

$$32t^{2}-26t-22=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 32 for a, -26 for b, and -22 for c in the quadratic formula.

$$t=\frac{-\left(-26\right)±\sqrt{\left(-26\right)^{2}-4\times 32\left(-22\right)}}{2\times 32}$$

Do the calculations.

$$t=\frac{26±6\sqrt{97}}{64}$$

Solve the equation t=\frac{26±6\sqrt{97}}{64} when ± is plus and when ± is minus.

$$t=\frac{3\sqrt{97}+13}{32} t=\frac{13-3\sqrt{97}}{32}$$

Since n=t^{2}, the solutions are obtained by evaluating n=±\sqrt{t} for positive t.

$$n=\frac{\sqrt{\frac{3\sqrt{97}+13}{8}}}{2} n=-\frac{\sqrt{\frac{3\sqrt{97}+13}{8}}}{2}$$